3.1-Quadratic Functions & Inequalities

Quadratic Functions: Quadratic functions are polynomial functions of the form cbxaxxf ++= 2)( . A quadratic function may

also be written in the form khxaxf +−= 2)()( .

Vertex Form of a Quadratic Function:

When a quadratic equation is written in the form khxaxf +−= 2)()( the vertex is very easy to determine. In fact, there really is no work at all because this form of a quadratic function gives us the vertex directly. In this form, the vertex is:

),(),( khyx = Example: Given the equation 5)4(2 2 +−= xy determine the vertex. Solution: The vertex is )5,4(),( =kh .

Example: Given the equation 4

5

3

23

2

+

+= xy determine the vertex.

Solution: The vertex is

−=4

5,

3

2),( kh .

Rewriting a Quadratic Function in Vertex Form: Since the vertex form of a quadratic equation is so handy, it would be nice to know how to write any equation in this form. All we need to do is complete the square on the quadratic equation in the form

cbxaxxf ++= 2)( to get it in the form khxaxf +−= 2)()( . Example: Write the equation 242 ++= xxy in vertex form and identify the vertex. Solution: To complete the square on this equation we will modify the procedure that we learned previously. This modification will allow us to complete the square while only working on the right side of

the function. As always, find the value of2

2

b. Then add and subtract this value from the right side of the

function. Also, group the terms of the trinomial square in parentheses.

24)44()(

24)(2

2

+−++=

++=

xxxf

xxxf

Now write the trinomial square in its factored form and simplify. The result is the equation in vertex form.

2)2()( 2 −+= xxf

The vertex is (-2,-2).

Example: Write the equation 562 −+= xxy in vertex form and identify the vertex. Solution: I will solve without explanation.

14)3(

59)96(

56

2

2

2

−+=−−++=

−+=

xy

xxy

xxy

The vertex is (-3,-14). Example: Write the equation 682 2 ++= xxy in vertex form and identify the vertex. Solution: As always, the “a” term must be equal to 1. Since we are working only on the right side, we must factor the 2 out on the right instead of dividing both sides of the equation by 2 as we did previously.

)34(2

6822

2

++=

++=

xxy

xxy

Now complete the square inside the parentheses.

2)2(2

)1)2((2

)34)44((2

)34(2

2

2

2

2

−+=−+=

+−++=

++=

xy

xy

xxy

xxy

The vertex is (-2,-2). Example: Write the equation 563 2 ++= xxy in vertex form and identify the vertex. Solution:

2)1(3

3

2)1(3

3

51)12(3

3

523

563

2

2

2

2

2

++=

++=

+−++=

++=

++=

xy

xy

xxy

xxy

xxy

The vertex is( )2,1− .

Graphing a Quadratic Function: Graphing a quadratic function in vertex form is straight forward since khxaxf +−= 2)()(

Is a transformation of the basic quadratic function 2)( xxf = . Example: Sketch the graph of the function 7)2()( 2 −−= xxf

Solution: The vertex is )7,2( −

Example: Sketch the graph of the function 4)5()( 2 ++−= xxf Solution: The vertex is )4,5(− . Notice that the “a” value in khxaxf +−= 2)()( is negative indicating a reflection over the x-axis.

Example: Sketch the graph of the function 342)( 2 +−−= xxxf Solution: Rewrite the function in the form khxaxf +−= 2)()(

( )

( )

5)1(2)(

32)1(2)(

31)1(2)(

3)112(2)(

322)(

342)(

2

2

2

2

2

2

++−=+++−=

+−+−=

+−++−=

++−=

+−−=

xxf

xxf

xxf

xxxf

xxxf

xxxf

Notice that the function is stretched by a factor of 2.

Characteristics of a Parabola:Parabolas have special characteristics that help us to sketch them. Each parabola has either a highest or lowest point called a vertex. A graph of a quadratic equation also has an vertical line that passes through the vertex. A parabola that opens upward has a lowest point with a minimum y-value occurring at the vertex. A parabola that opens downward has a highest point with a maximum y-value occurring at the vertex. No

Finding the Vertex of a ParabolaThe vertex of a parabola is the turning point of the parabola. There are two methods of finding the vertex. The first method is to use the vertex form of a quadratic equation. formula. The Vertex Formula:

The vertex is a point on the graph of a parabola pair. As an ordered pair the vertex may be written as:

Example: Given the parabola =y Solution: To find the x-value of the vertex

To find the y-value of the vertex, we substitute the xfor y.

Therefore, the vertex is )3,2( −− .

Characteristics of a Parabola: Parabolas have special characteristics that help us to sketch them. Each parabola has either a highest or

. A graph of a quadratic equation also has an axis of symmetryvertical line that passes through the vertex. A parabola that opens upward has a lowest point with a

value occurring at the vertex. A parabola that opens downward has a highest point with a value occurring at the vertex. Note that a parabola extends infinitely far

A parabola may open upward or downward. A parabola that opens upward has a leading term that is positive. A parabola that opens downward has a leading term that is negative.

423 2 −+= xxy Opens Up

142 2 ++−= xxy Opens Down

Finding the Vertex of a Parabola The vertex of a parabola is the turning point of the parabola. There are two methods of finding the vertex. The first method is to use the vertex form of a quadratic equation. The second method is to use the vertex

The vertex is a point on the graph of a parabola cbxaxy ++= 2 and may be described as an ordered pair. As an ordered pair the vertex may be written as:

( ) =yx,

−−a

bf

a

b

2,

2

582 2 ++ xx find the vertex.

value of the vertex use the vertex formula.

24

8

)2(2

8

2−=−=−=−=

a

bx

value of the vertex, we substitute the x-value back into the quadratic equation and evaluate

3

5168

5)2(8)2(2 2

−=+−=

+−+−=

y

y

y

Parabolas have special characteristics that help us to sketch them. Each parabola has either a highest or axis of symmetry, that is, a

vertical line that passes through the vertex. A parabola that opens upward has a lowest point with a value occurring at the vertex. A parabola that opens downward has a highest point with a

extends infinitely far in two directions.

A parabola may open upward or downward. A parabola that opens upward has a leading term that is positive. A parabola that opens downward has a leading term that is negative.

The vertex of a parabola is the turning point of the parabola. There are two methods of finding the vertex. The second method is to use the vertex

and may be described as an ordered

value back into the quadratic equation and evaluate

Example: Given the parabola 83182 +−= xxy find the vertex. Solution: To find the x-value of the vertex use the vertex formula.

92

18

)1(2

)18(

2==−−=−=

a

bx

To find the y-value of the vertex, we substitute the x-value back into the quadratic equation and evaluate for y.

2

8316281

83)9(18)9( 2

=+−=

+−=

y

y

y

Therefore, the vertex is )2,9( .

Axis of Symmetry: The axis of symmetry of a parabola is the imaginary line that divides the parabola into two equal halves. Each half of the parabola is a reflection of the other half over the axis of symmetry. The axis of symmetry is a vertical line and will consequently have an equation of the form ax = as do all vertical lines. The vertex will always be on the axis of symmetry. Consequently, the x-value of the vertex will also be the axis of symmetry. Example: Find the vertex and axis of symmetry of 7)2( 2 −−= xy Solution: The vertex is (2, -7). The axis of symmetry is2=x . Example: Find the vertex and axis of symmetry of 4)5( 2 −+= xy Solution: The vertex is (-5, -4). The axis of symmetry is 5−=x .

Extreme Value of a Parabola: Because the vertex is the turning point of the parabola, it also represents the maximum or minimum point of the parabola. This value is also called the extreme value of the parabola. Upward Opening Parabola A parabola opens upward if its leading term is positive. If a parabola opens upward, its extreme value will be a minimum value. The minimum value is the y-value of the vertex. Example: Find the extreme value of the parabola 7)2( 2 −−= xy and determine if it opens upward or downward. Solution: Since the leading term is positive the extreme value is a minimum. The minimum value is y = -7

Downward Opening Parabola A parabola opens downward if its leading term is negative. If a parabola opens downward, its extreme value will be a maximum point. The maximum value is the y-value of the vertex.

Example: Find the extreme value of the parabola 6)3(5 2 ++−= xy and determine if it opens upward or downward. Solution: Since the leading term is negative the extreme value is a maximum. The maximum value is y = 6

Intercepts of a Quadratic Equation:The intercepts of a quadratic equation are The procedure for finding the intercepts is the same as for linear equations (or any equations for that matter). To find x-intercepts: Let y = 0 and solve for x. This is what we are doinsolve a quadratic equation….finding the xintercepts should always be expressed as ordered pairs. There may be as many as two x-intercepts. The discriminant helps us determine how many solutions or x

Intercepts of a Quadratic Equation: The intercepts of a quadratic equation are where the graph of the parabola crosses the xThe procedure for finding the intercepts is the same as for linear equations (or any equations for that

Let y = 0 and solve for x. This is what we are doing when we solve a quadratic equation….finding the x-intercepts. The x-intercepts should always be expressed as ordered pairs. There

intercepts.

The discriminant helps us determine how many solutions or x-intercepts the equation will have.

To find the y-intercept: Let x = 0 and solve for y. The y-intercept should always be expressed as an ordered pair. There will only be one yintercept.

where the graph of the parabola crosses the x-axis and the y-axis. The procedure for finding the intercepts is the same as for linear equations (or any equations for that

uation will have.

intercept should always be expressed as an ordered pair. There will only be one y-

Example: Find the intercepts of the equation 1582 ++= xxy . Solution: To find the x-intercepts, let y = 0 and solve for x.

5,3

)5)(3(0

1580

1582

2

−=−=++=

++=++=

xx

xx

xx

xxy

The x intercepts are (-3, 0) and (-5, 0)

To find the y-intercept, let x = 0 and solve for y.

15

15)0(80

1582

2

=++=

++=

y

y

xxy

The y-intercept is (0, 15)

Example: Find the intercepts of the equation 32 2 −−= xxy . Solution: To find the x-intercepts, let y = 0 and solve for x.

2

3,1

)1)(32(0

320

322

2

=−=

+−=−−=−−=

xx

xx

xx

xxy

The x intercepts are (-1, 0) and

0,

2

3

To find the y-intercept, let x = 0 and solve for y.

3

30)0(2

322

2

−=−−=

−−=

y

y

xxy

The y-intercept is (0, -3)

Example: Find the intercepts of the equation. 284 2 −= xy Solution: To find the x-intercepts, let y = 0 and solve for x.

7

284

2840

284

2

2

2

±=

=

−=

−=

x

x

x

xy

The x intercepts are ( )0,7 and ( )0,7−

To find the y-intercept, let x = 0 and solve for y.

28

28)0(4

2842

2

−=−=

−=

y

y

xy

The y-intercept is (0, -28)

Quadratic Inequalities A quadratic inequality in one variable is an inequality that contains a quadratic expression. The standard form of a quadratic inequality 2 ++ bxaxwhere any of the inequality signs <, Example: The following are all quadratic inequalities.

2 +x

4 −x

Solutions of a Quadratic Inequality:Solving a quadratic inequality is very similar to solving a quadratic equation. Remember that the solution to a quadratic equation 2 ++ cbxax

Procedure for Solving a Quadratic Inequality:Step 1: Solve the related quadratic equation using any appropriate method. Step 2: Use the solutions from Step 1 to divide the number line into either two or three intervals as necessary. Step 3: Select a Test Point in each interval and determine its sign value. Step 4: The solution will consist of the intervals with the

A quadratic inequality in one variable is an inequality that contains a quadratic expression. The standard

0<+ c looks very much like the standard form of a quadratic equation ≥≤> ,,, may be used in place of an equal sign.

The following are all quadratic inequalities. 943 <−+ x xx 243 2 >−

13 2 ≥− x 324 2 −≤−− xxx

Solutions of a Quadratic Inequality: Solving a quadratic inequality is very similar to solving a quadratic equation. Remember that the solution

0=c represents the values where the parabola crosses the x

The solution to the quadratic inequality

2 <++ cbxaxwhere the parabola is below the x The solution to the quadratic inequality

2 >++ cbxaxwhere the parabola is above the x

Procedure for Solving a Quadratic Inequality: Solve the related quadratic equation using any appropriate method.

Use the solutions from Step 1 to divide the number line into either two or three intervals as necessary.

Select a Test Point in each interval and determine its sign value.

The solution will consist of the intervals with the appropriate sign value.

A quadratic inequality in one variable is an inequality that contains a quadratic expression. The standard

looks very much like the standard form of a quadratic equation may be used in place of an equal sign.

4

Solving a quadratic inequality is very similar to solving a quadratic equation. Remember that the solution where the parabola crosses the x-axis.

The solution to the quadratic inequality 0 represents the values

the parabola is below the x-axis.

The solution to the quadratic inequality 0 represents the values

the parabola is above the x-axis.

appropriate sign value.

Example: Solve the quadratic inequality Solution: Step 1: Solve the related equationx

Step 2: The values x = -5 and x = 2 are called equal to 0. We will use the boundary values to divide the number line into three intervals. These boundary values will be the endpoints of the intervals. The three intervals are Step 3: Select a test point in each of these intervals and evaluate this point in the original quadratic

1032 −+ xx . Interval Test Point Evaluate in Quadratic

)5,( −−∞ -6 )6( 2−)2,5(− 0 )0( 2 +

(2, )∞ 3 )3( 2 +

Solve the quadratic inequality 01032 <−+ xx .

01032 =−+ xx .

2,5

0)2)(5(

01032

=−==−+

=−+

xx

xx

xx

5 and x = 2 are called boundary values and they represent where the quadratic is equal to 0. We will use the boundary values to divide the number line into three intervals. These boundary values will be the endpoints of the intervals. The three intervals are ,5(),5,( −−−∞

Select a test point in each of these intervals and evaluate this point in the original quadratic

Evaluate in Quadratic Sign Value 810)6(32 =−−+ 8>0 or positive

1010)0(3 −=−+ -10<0 or negative

810)3(3 =−+ 8>0 or positive

Step 4: Since we are interested in knowing where the quadratic is <interval where the sign value is negative or <0. Consequently the solution to the quadratic inequality is )2,5(− .

and they represent where the quadratic is equal to 0. We will use the boundary values to divide the number line into three intervals. These boundary

),2(),2, ∞ .

Select a test point in each of these intervals and evaluate this point in the original quadratic

Since we are interested in knowing where the quadratic is <0 the solution will be the

where the sign value is negative or <0.

Consequently the solution to the quadratic

Example: Solve the quadratic inequality Solution: Step 1: Rewrite inequality with a zero on the right side and then solve the related equation.

Step 2: The values x = 5 and x = 1 are the boundary values and they represent where the quadratic is equal to 0. We will use the boundary values to divide the number line into three intervals. These boundary values will be the endpoints of the intervals. The three intervals aresign the boundary values may be part of the solution and we use brackets in the interval notation. Step 3: Select a test point in each of these intervals and evaluate this point in the original quadratic

562 +− xx . Interval Test Point Evalua

]1,(−∞ 0 )0( 2 −]5,1[ 2 )2( 2 −),5[ ∞ 6 )6( 2 −

Solve the quadratic inequality 7262 −≥−− xx .

Rewrite inequality with a zero on the right side and then solve the related equation.

1,5

0)1)(5(

056

056

726

2

2

2

===−−

=+−≥+−

−≥−−

xx

xx

xx

xx

xx

The values x = 5 and x = 1 are the boundary values and they represent where the quadratic is equal use the boundary values to divide the number line into three intervals. These boundary values

will be the endpoints of the intervals. The three intervals are ),5[],5,1[],1,( ∞−∞ . Note that because of the sign the boundary values may be part of the solution and we use brackets in the interval notation.

Select a test point in each of these intervals and evaluate this point in the original quadratic

Evaluate in Quadratic Sign Value 55)0(6 =+− 5>0 or positive

35)2(6 −=+− -3<0 or negative

55)6(6 =+− 5>0 or positive

Step 4: Since we are interested in knowing where the quadratic is the interval where the sign value is positive or >0. Consequently the solution to the quadratic inequality is ,5[],1,(−∞

Rewrite inequality with a zero on the right side and then solve the related equation.

The values x = 5 and x = 1 are the boundary values and they represent where the quadratic is equal use the boundary values to divide the number line into three intervals. These boundary values

. Note that because of the ≥ sign the boundary values may be part of the solution and we use brackets in the interval notation.

Select a test point in each of these intervals and evaluate this point in the original quadratic

Since we are interested in knowing where the quadratic is 0≥ the solution will be the interval where the sign value is positive or >0. Consequently the solution to the quadratic

),∞ .

Example: Solve the quadratic inequality Solution: Step 1: Solve the related equation x

ix

x

x

x

5

25

25

025

2

2

2

±=−±=

−=

=+

Step 2: because the boundary values are imaginary, the parabola does not cross the x-axis. Consequently, the solution is either ),( ∞−∞ or no solution. Because the leading term is positive the parabola is an upturning parabola and must therefore be above the xinequality is looking for points where the quadratic is below the x-axis, there is no solution. Example: Solve the quadratic inequality Solution: Step 1: Solve the related equation −

ix

x

x

x

6

36

36

036

2

2

2

±=−±=

−==−−

Step 2: because the boundary values are imaginary, the parabola does not cross the x-axis. Consequently, the solution is either ),( ∞−∞ or no solution. Because the leading term is negative the parabola is a down turning parabola and must therefore be below the x-axis. Since the inequality is looking for points where the quadratic is below the xthe solution is ),( ∞−∞ .

Solve the quadratic inequality 0252 ≤+x .

0252 =+x

because the boundary values are imaginary, the axis. Consequently, the

or no solution. Because the leading term is positive the parabola is an upturning

above the x-axis. Since the inequality is looking for points where the quadratic is

axis, there is no solution.

Solve the quadratic inequality 036 2 <−− x .

036 2 =−− x

because the boundary values are imaginary, the axis. Consequently, the

or no solution. Because the leading parabola is a down turning parabola and

axis. Since the inequality is looking for points where the quadratic is below the x-axis,

3.1-Applications

Example: Mike wants to enclose a rectangular area for his rabbits alongside his large barn using 30 feet of fencing. What dimensions will maximize the area fenced in if the barn is used for one side of the rectangle? What is the maximum area? Solution: This problem involves two geometry formulas; P=2L+2W and A+LW. Begin with the perimeter formula by letting x = the length of the rectangular area. Then the width may be determined as follows:

xw

wx

wLP

230

230

2

−=+=+=

In the above formula only one width is required since the barn occupies one side. Now substitute the value of length and width into the area formula.

2230

)230(

xxA

xxA

LWA

−=

−==

To maximize this we need to find the vertex:

( ) =yx,

−−a

bf

a

b

2,

2

Use a = -2, and b = 30

5.74

30 =−−=x

Evaluate the function at x = 7.5

5.112)5.7(

)5.7(2)5.7(30)5.7(

230)(2

2

==−==

−==

Ay

Ay

xxxAy

The maximum area is 112.5 sq ft and will have a length of 7.5 ft and a width of 15.

Example: If an archer shoots an arrow straight upward with an initial velocity of 160 ft/sec from a height of 8 ft., then its height above the ground in feet at time t in seconds is given by the function

816016)( 2 ++−= ttth . What is the maximum height reached by the arrow? How long does it take to reach the maximum height? How long does it take for the arrow to reach the ground? Solution: The maximum height will be the vertex of the parabola. Since the leading term is negative, the parabola will be down turning and so this makes sense. The vertex is a point on the graph of a parabola and may therefore be described as an ordered pair. As an ordered pair the vertex may be written as:

( ) =yx,

−−a

bf

a

b

2,

2

Use the values a = -16, and b = 160 and substitute into the formula to find the x-value of the vertex.

5)16(2

160 =−

−=x

Now find the function value at x = 5.

408)5(

8800400)5(

8)5(160)5(16)5(

816016)(2

2

=++−=

++−=++−=

h

h

h

ttth

The maximum height on the graph is the y-value of the vertex and is therefore 408 ft. The time to reach the maximum height is the x-value of the vertex. It takes 5 seconds to reach the maximum height. To determine when the arrow hits the ground we need to determine the time when the height is zero. Therefore, let h (t) =0 and solve 816016)( 2 ++−= ttth for t.

12020

8160160

816016)(

2

2

2

−−=

++−=++−=

tt

tt

ttth

Solving this equations gives the solutions:

05.10

05.552

515

=±=

±=

t

t

t

It will take 10.05 seconds for the arrow to hit the ground. The other solution would be -0.05 which of course makes mathematical sense but not in the context of the problem.

Example: Flying too fast or too slow wastes fuel. For the Spirit of St. Louis miles per pound of fuel M was a function of airspeed A in miles per hour, modeled by the formula

.01.5127.02000653.0 −+−= AAM At what airspeed does the Spirit of St. Louis obtain its best fuel burn (greatest number of miles per pound of fuel)? What is the greatest fuel burn? Solution: Notice that the function is a quadratic that expresses fuel burn M as a function of airspeed A. Consequently, we need to maximize the fuel burn by finding the vertex of the parabola. Since rewriting this function in vertex form would be an unpleasant affair, I will use the vertex formula.

( ) =yx,

−−a

bf

a

b

2,

2

Use a = -0.000653, b = 0.127.

24.97)000653.0(2

127.0 =−−=x

Evaluate the function at x = 194.5

17.1)24.97(

.01.5)24.97(127.0)24.97(000653.0)24.97( 2

=−+−=

M

M

The maximum fuel burn of 1.17 miles per pound of fuel is obtained at 97.24 miles per hour.

Example: If a football is kicked straight up with an initial velocity of 128 ft/sec from a height of 5 ft, then its height above the earth is a function of time given by 512816)( 2 ++−= ttth . At what time does the ball reach this height? What is the maximum height reached by the ball? At what time does the ball hit the ground? Solution: Find the vertex of the parabola in the form (t, h (t)) using the given function. The time that the ball reaches its maximum height is given by the t-value of the vertex and the maximum height of the ball is given by the h(t) value of the vertex.

432

128

2=

−−=−=

a

bt

Use this value to find h(t).

261)4(

5512256)4(

5)4(128)4(16)4(

512816)(2

2

=++−=

++−=++−=

h

h

h

ttth

The ball reaches a maximum height of 261 ft at 4 seconds after being kicked. To find the time that the ball hits the ground we must solve the equation

5128160 2 ++−= tt Using the quadratic equation we obtain

03.8

039.032

24.129128

=−=

−±−=

t

t

t

Therefore, the ball hit the ground approximately 8.03 seconds after being kicked.

Example: An air charter company’s weekly revenue in dollars is given by xxxR 50005)( 2 +−= where x is the number of hours flown during a month. 1. What is the maximum revenue? 2. For what number of hours is the revenue maximized? 3. On what interval is )(xR increasing? Decreasing? 4. Over what interval is 0)( >xR ? Solution: Find the vertex of the parabola in the form (x, R(x)) using the given function.

50010

5000

2=

−−=−=

a

bx

Use this value to find R(x).

000,250,1)500(

000,500,2000,250,1)500(

)500(5000)500(5)500(

50005)(2

2

=+−=

+−=+−=

R

R

R

xxxR

The maximum revenue is $1,250,000 which is earned at 500 hours flight time per month. Because the parabola is down turning, the function is increasing over the interval )500,0( and decreasing over the interval ).,500( ∞ Consequently, it is to the company’s advantage to fly as close to 500 hours per month as possible. To find where 0)( >xR we must first find the x-intercepts by solving the equation

1000

0

)1000(50

500050 2

==

−−=+−=

x

x

xx

xx

Because this is a down turning parabola, 0)( >xR over the interval (0, 1,000).

Example: Eagle View tour company gives a jeep tour of the south rim of the Grand Canyon to one person for $78.00 To increase their business, the company has offered Westwind Air Tours a special rate that would lower the price by $2 per person for each additional person, up to 30 people.

1. Write the price per person p as a function of the number of people. 2. Write the revenue as a function of the number of people on the tour. 3. What is the maximum revenue for the tour?

Solution: The price per person is given by the function. To find the constant, add the discount (2) to the normal rate (78). Then, subtract the discount times the number of people (x).

xxP 280)( −= The revenue function is the price per person P(x) times the number of people x.

2280)(

)280()(

)()(

xxxR

xxxR

xxPxR

−=

−==

The maximum revenue is found by finding the vertex of the parabola of the revenue function.

204

80

2=

−−=−=

a

bx

Use this value to find R(x).

800)20(

8001600)20(

)20(2)20(80)20(

280)(2

2

=−=

−=−=

R

R

R

xxxR

The maximum revenue is $800.00 which is earned when 20 people are on the tour.

Example: A soldier in basic training fires a rocket propelled grenade straight up from ground level with an initial

velocity of 256 ft/sec. Find the maximum height reached by the RPG. How long does it take the RPG to reach its

maximum height? How long does it take to reach the ground?

Solution: The maximum height will be the vertex of the parabola. Since the leading term is negative, the parabola

will be down turning and so this makes sense. The function that models this situation is ttth 25616)( 2 +−= . I

will use the vertex form of the equation to determine the vertex.

1024)8(16)(

)646416(16)(

)16(16)(

25616)(

2

2

2

2

+−−=−+−−=

−−=

+−=

tth

ttth

ttth

ttth

The vertex of the parabola is (8, 1024). Therefore, the RPG reaches its maximum height of 1024 ft. in 8 sec.

To find the time it takes to reach the ground we need to solve the equation.

16,0

0)16(16

256160

25616)(2

2

===−−

+−=+−=

tt

tt

tt

ttth

The RPG reaches the ground in 16 seconds. Notice that it was also on the ground at zero seconds as would be

expected.

Example: A company’s weekly revenue in dollars is given by

222000)( xxxR −= , where x is the number of items

produced during a week.

A) For what values is R(x) > 0?

B) For what number of items is revenue maximized? What is the maximum revenue?

C) On what interval is R(x) increasing? Decreasing?

Solution: To find where R(x)>0 we need to solve the inequality 022000 2 >− xx .

First solve the corresponding equation to find the x-intercepts.

000,1,0

0)1000(2

022000 2

===+−−

=−

xx

xx

xx

Because this is a down-turning parabola, the graph will be above the x-axis between these values, R(x)>0 over the

interval (0, 1,000).

A) To find the number of items for which revenue is maximized we need to find the vertex.

000,500)500(2)(

000,500)000,2501000(2)(

)000,250000,2501000(2)(

)1000(2)(

22000)(

2

2

2

2

2

+−−=++−−=

−+−−=−−=−=

xxR

xxxR

xxxR

xxxR

xxxR

The number of items for which revenue is maximized is 500 and the revenue on that number will be $500,000.

B) Since the vertex is (500, 500,000) and the parabola is down turning, the function will be increasing from

)500,(−∞ and decreasing from ),500( ∞